Question

Cyanide recovered from the refining of gold ore
can be determined indirectly by EDTA titration. A known
excess of Ni2 is added to the cyanide to form tetracyanonickelate(
II):
4CN + Ni ---> Ni(CN)4
When the excess Ni2 is titrated with standard EDTA,
Ni(CN)4 does not react. In a cyanide analysis, 12.7 mL of
cyanide solution were treated with 25.0 mL of standard solution
containing excess Ni to form tetracyanonickelate. The excess
Ni required 10.1 mL of 0.013 0 M EDTA for complete
reaction. In a separate experiment, 39.3 mL of 0.013 0 M EDTA
were required to react with 30.0 mL of the standard Ni
solution. Calculate the molarity of CN in the 12.7-mL sample
of unknown.

Answer 1

Ni²⁺ + EDTA^4- -----> Ni-EDTA ^2-

Number of moles of EDTA reacted ,n = Molarity x volume in L

                                                     = 0.0130 M x 39.3 mL x 10^-3 L/mL

                                                    = 5.11x10^-4 mol

From the above reaction ,

1 mole of Ni²⁺ reacts with 1 mole of EDTA

5.11x10^-4 mol of Ni2+ reacts with 5.11x10^-4 mol of EDTA

So concentration of Ni²⁺ is = number of moles / volume in L

                                    = 5.11x10^-4 mol / (30.0x10^-3 L)

                                    = 0.017 M

Therefore the number of moles of Ni²⁺ reacted with EDTA is = number of moles of EDTA

                                                                                       = Molarity x volume in L

                                                                                       = 0.013 M x 10.1 mL x 10^-3 L/mL

                                                                                       = 1.313x10^-4 moles = n

The number of moles of Ni2+ added to CN- is , n' = Molarity x volume in L

                                                                       = 0.017 M x 25.0x10^-3 L

                                                                       = 4.25x10^-4 mol

So the actual number of moles of Ni+ reacted with CN- is , N = n' - n = 2.94x10^-4 mol

4CN^- + Ni ---> Ni(CN)₄

From the above balanced reaction ,

1 mole of Ni reacts with 4 moles of CN-

2.94x10^-4 mol of Ni reacts with 4x2.94x10^-4 mol= 1.17x10^-3 mol of CN-

So Molarity of CN , M = number of moles / volume in L

                                = (1.17x10^-3 mol) / (12.7 mLx10^-3L/mL)

                               = 0.0925 M

Therefore the molarity of CN in the sample is 0.0925 M